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Ratling [72]
2 years ago
12

What is the approximate percentage of nitrogen in the Earth's current atmosphere?

Chemistry
1 answer:
RoseWind [281]2 years ago
7 0

Answer:

78% nitrogen

Explanation:

It has been long agreed upon that our atmosphere is highly abundant with nitrogen. Our atmosphere is composed of  78% nitrogen and then other gases follow such as 21% oxygen, 0.9% argon, and 0.03% carbon dioxide with very small percentages of other elements.

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A 25.225 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 75.815 g of wa
Rudik [331]

Answer:

1.43 (w/w %)

Explanation:

HCl reacts with NH3 as follows:

HCl + NH3 → NH4+ + Cl-

<em>1 mole of HCl reacts per mole of ammonia.</em>

Mass of NH3 is obtained as follows:

<em>Moles HCl:</em>

0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>

<em>Mass NH3 in the aliquot:</em>

3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.

Mass of sample + water = 22.225g + 75.815g = 98.04g

Dilution factor: 98.04g / 14.842g = 6.6056

That means mass of NH3 in the sample is:

0.0545g * 6.6056 = 0.36g NH3

Weight percent is:

0.36g NH3 / 25.225g * 100

<h3>1.43 (w/w %)</h3>
6 0
3 years ago
An alkane with the formula C6H14 can be prepared by hydrogenation of either of two precursor alkenes having the formula C6H12.
ankoles [38]
Two precursor alkenes

    H₃C  CH₃
         I   I
H₂C=C-CH-CH₃    2,3-dimethyl-1-butene

   H₃C      CH₃
        I       I
H₃C-CH=CH-CH₃    2,3-dimethyl-2-butene


alkane

   H₃C     CH₃
        I      I
H₃C-CH-CH-CH₃    2,3-dimethylbutane

    H₃C  CH₃                        H₃C     CH₃
         I   I                                   I      I
H₂C=C-CH-CH₃  + H₂ → H₃C-CH-CH-CH₃ 
  
    H₃C   CH₃                       H₃C     CH₃
        I    I                                   I      I
H₂C-C=CH-CH₃  + H₂ → H₃C-CH-CH-CH₃ 

6 0
3 years ago
Read 2 more answers
What is the element present for copper
amm1812

Copper is an brown-orange color which it's atomic number is 29. With high thermal and electricity conductivity with it's smooth surface.

8 0
3 years ago
A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is
DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

  • C = concentration of acid (HCl) = 0.0470 M,
  • V = volume of acid = 16.0 mL,
  • C' = concentration of base (ammonia solution) and
  • V' = volume of base = 26.0 mL.
<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

Substituting the values of the variables into the equation, we have

C' = CV/V'

C' = 0.0470 M × 16.0 mL/26.0 mL

C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] =  where

  • [H⁺] = concentration of HCl at equivalence point,
  • [OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
  • Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

[H⁺] = 34.95 × 10⁻¹⁴

[H⁺] = 3.495 × 10⁻¹³

<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

pH = 13 - 0.5434

pH = 12.4566

pH ≅ 12.46

So, the pH at equivalence point is 12.46

Learn more about pH at equivalence point here:

brainly.com/question/25487920

3 0
2 years ago
How many grams are in 1.3 moles of Cr?
Verizon [17]

Answer:

40.38 grams

Explanation:

6 0
2 years ago
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